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2 years ago

Find the perimeter and area of the shape.

Mathematics

2 answers:

belka [17]2 years ago

7 0

Answer:

The perimeter is 41

Step-by-step explanation:

The perimeter of a shape is just the sum of all its sides added up. In this case, that would mean the perimeter is:

12 + 9 + 9 + 4.6 + 6.4 = P
30 + 11 = P
41 = P

Hope this helps!

STatiana [176]2 years ago

4 0

The answer to the question is 41

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If f(x, y) = esin(x + y) and D = [−π, π] × [−π, π], show the following. 1 e ≤ 1 4π2 D f(x, y) dA ≤ e The area of the region D is

Digiron [165]

I'm guessing the purpose of this exercise is to find the average value of $f(x,y)=e^{\sin(x+y)}$ over the region $D$ ,

$D=\left\{(x,y)\mid (x,y)\in[-\pi,\pi]^2\right\}$

which is

$\displaystyle\frac{\displaystyle\iint_Df(x,y)\,\mathrm dx\,\mathrm dy}{\displaystyle\iint_D\mathrm dx\,\mathrm dy}$

The denominator is the measure (area) of $D$ , which is easy to compute:

$\displaystyle\iint_D\mathrm dx\,\mathrm dy=\int_{-\pi}^\pi\int_{-\pi}^\pi\mathrm dx\,\mathrm dy=(2\pi)^2=4\pi^2$

Not to be confused with the integral in the numerator:

$\displaystyle\iint_Df(x,y)\,\mathrm dx\,\mathrm dy=\int_{-\pi}^\pi\int_{-\pi}^\pi e^{\sin(x+y)}\,\mathrm dx\,\mathrm dy$

but this integral is difficult to compute, hence the inequalities. We have

$-1\le\sin(x+y)\le1$

$\implies\dfrac1e\le e^{\sin(x+y)}\le e$

$\implies\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}e\le\iint_De^{\sin(x+y)}\,\mathrm dx\,\mathrm dy\le\iint_De\,\mathrm dx\,\mathrm dy$

$\implies\displaystyle\frac{4\pi^2}e\le\iint_De^{\sin(x+y)}\,\mathrm dx\,\mathrm dy\le4\pi^2e$

3 0

3 years ago

Hello :) <br> Please help me solve this <br><br> Explain answer

svet-max [94.6K]

Answer:

$Scale\ Factor = \frac{1}{2}$

ABC dilated

Step-by-step explanation:

Given

Triangles ABC and DEF

Required

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First, we pick a side on ABC

$AB = 6$

Pick a corresponding side on DEF

$DE = 3$

The scale factor is then calculated as:

$Scale\ Factor = \frac{DE}{AB}$

$Scale\ Factor = \frac{1}{2}$

i.e. ABC was dilated by 1/2

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2 years ago

Which of the relations given by the following sets of ordered pairs is a function?

solniwko [45]

Answer:

Step-by-step explanation:

You can solve this by using the vertical line test. When you do the vertical line test, the vertical line should only pass through one point on the function. That means that there can to be only one value of x for every y. Set B is the only set where the x value doesn't repeat.

8 0

3 years ago

RCX)=2VX SCx) = x(RSX4=

aev [14]

Given the two functions:

$\begin{gathered} R(x)=2\sqrt[]{x} \\ S(x)=\sqrt[]{x} \end{gathered}$

We need to find (RoS)(4). THis is the functional composition. We take S(x) and put it into R(x) and then put "4" into that composed function. Shown below is the process:

$(RoS)(x)=2\sqrt[]{\sqrt[]{x}}$

When we plug in "4", into "x", we have:

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1 year ago

A+5=5a+5 how do i do this problem

Natali5045456 [20]

Answer:

0 = a

Step-by-step explanation:

The first thing to do is subtract 5 from both sides.

a + 5 = 5a + 5

- 5 - 5

The 5's cancel out.

Now we have a + 0 = 5a

You next subtract the a from both sides.

a + 0 = 5a

-a -a

0 = 4a

Lastly would to get a by itself so then we divide 4 from both sides

0 = 4a

_ _

4 4

0 = a

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2 years ago